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A useful short-interval approximation replaces a battery by an ideal electromotive force $\mathcal{E}$ in series with a positive internal electrical resistance $r$. This is the Thévenin terminal response with open-circuit voltage $\mathcal{E}$ and Thévenin resistance $r$. The model describes the terminals without asserting that there is a separate physical resistor hidden inside the cell. Both $\mathcal{E}$ and $r$ can change with composition, temperature, electric current and history.^[ref]

For a resistive electrical load $R>0$, take $I>0$ to mean discharge: conventional current leaves the positive terminal. Kirchhoff's voltage law and Ohm's law giveFor the stipulated $\mathcal{E}=3\mathrm{V}$, $r=1\Omega$, $R=4\Omega$,An open-circuit voltage of $3\mathrm{V}$ therefore does not promise $3\mathrm{V}$ at this electrical load. Increasing the load electrical resistance reduces electric current and brings $V_L$ closer to $\mathcal{E}$, as in Thévenin load response.

The signed relation also explains charging within this model. If an external source drives electric current into the positive terminal, $I<0$ and $V_L=\mathcal{E}+|I|r$. The terminal electric potential difference then exceeds $\mathcal{E}$. This algebra is not evidence that every battery is rechargeable.^[ref]

A prescribed voltage source or current source used to probe a terminal response. For a suppressed linear resistive circuit, the ratio of terminal voltage to entering electric current determines the Thévenin resistance when finite.

In a Wheatstone bridge, the galvanometer connects the two comparison junctions. Its purpose is to locate a null measurement, rather than to measure the branch electric current through the unknown resistor. Using Wheatstone bridge detector loading and Thévenin's theorem, a settled moving-coil galvanometer with winding electrical resistance $R_g$ obeysHere $V_{\mathrm{oc}}$ is the bridge's open-circuit voltage and $R_{\mathrm{th}}$ its Thévenin resistance. For a functioning passive detector with finite positive electrical resistances, exact $I_g=0$ implies $V_{\mathrm{oc}}=0$, preserving Wheatstone bridge balance. Away from balance, the detector loads the junctions; its electrical resistance cannot be ignored.

For the illustrative $R_g=100\Omega$, $S_I=3000rad/A$ and a chosen $R_{\mathrm{th}}=1900\Omega$, a $1.00\mathrm{mV}$ bridge open-circuit voltage produces $0.500\mu \mathrm{A}$ and $0.0859^{\circ }$. That lies inside the stipulated $0.10^{\circ }$ zero tolerance. The corresponding resolvable open-circuit voltage scale is about $1.16\mathrm{mV}$, not arbitrarily small. Actual limits also depend on the zero stability and calibration.

Use the detector checks in the slide-wire resistance bridge experiment: perturb the comparison to either side of the proposed balance and require opposite signs, while confirming the supply is on. A broken connection, a jammed pointer or an inactive supply can also give a zero reading. Conversely, a healthy bridge detector reads zero while the bridge arms still carry electric current. A zero deflection alone proves neither a broken conducting path nor infinite sensitivity.

A negative ratio of terminal voltage to entering electric current, or a negative differential slope in a more general terminal relation. A linear dependent electrical source can produce a negative Thévenin resistance without any of the network's ordinary resistors having a negative value.

Resistor tolerance specifies the permissible manufacturing deviation of a resistor's electrical resistance from its nominal value under stated conditions. For nominal electrical resistance $R_{\mathrm{nom}}>0$ and fractional resistor tolerance $0\le \tau <1$, the specified range is $R_{\mathrm{nom}}(1-\tau )\le R\le R_{\mathrm{nom}}(1+\tau )$. This differs from measurement uncertainty: resistor tolerance describes a component specification, while measurement uncertainty describes knowledge of a measured value; the tolerance alone is not a probability distribution for that value.

Two nominally equal resistors can have different actual electrical resistances, changing their voltage divider ratio even without an external electrical load. In a loaded voltage divider, use the actual electrical resistances to calculate both the unloaded output and the Thévenin resistance governing voltage divider loading error. A mismatch from resistor tolerance and a change from voltmeter loading are distinct effects that can occur together.

In the pictured original linear resistive circuit, the ideal voltage source keeps its upper node at $12\mathrm{V}$ relative to $b$. Let $R_1=2\mathrm{k\Omega }$ connect that node to $a$, and $R_2=4\mathrm{k\Omega }$ connect $a$ to $b$. With the external electrical load removed, the voltage divider givesTo find the Thévenin resistance, Thévenin source suppression replaces the ideal $12\mathrm{V}$ voltage source by a wire. Both resistors now join $a$ to $b$ in parallel, givingThis is the electrical resistance looking into the original network with its independent electrical source suppressed; the external electrical load is excluded.

An independent check uses Kirchhoff's current law at $a$, without suppressing anything:The Thévenin terminal response thus follows directly from the original topology. Shorting $a$ to $b$ in this ideal calculation gives $I_{\mathrm{sc}}=6\mathrm{mA}$: the lower resistor has zero voltage and the upper resistor limits the electric current. All numerical values here and in the figures are stipulated model values, not laboratory data.

An authored teaching experiment uses an adjustable steady-voltage supply, a digital multimeter, fixed resistors, and a $10\mathrm{k\Omega }$ potentiometer or resistance decade box. Its original circuit uses a $10\mathrm{V}$ supply and $3.3$, $6.8$, and $4.7\mathrm{k\Omega }$ resistors; the student compares $8.2$ and $2.2\mathrm{k\Omega }$ electrical loads. The readout is load voltage, recorded against prediction and deviation. The student then removes the electrical load to measure the open-circuit voltage, disconnects the supply and substitutes a wire to measure the Thévenin resistance with the multimeter's resistance mode, and restores the supply for an independent method: adjust the variable electrical load until the voltmeter reads half the unloaded voltage. Finally the student builds the equivalent and repeats both load measurements. Its tables are blank recording forms, not reported observations. ^[ref]

The half-voltage inference assumes negligible voltmeter loading and positive $R_{\mathrm{th}}$. A finite voltmeter input electrical resistance $R_m$ actually gives an apparent unloaded readingWhen the same voltmeter remains connected across a variable electrical load $R_L$, the network sees $R_p=R_L{R}_m/(R_L+R_m)$. Setting its reading to $V_m/2$, rather than to the true $V_{\mathrm{oc}}/2$, givesThus the uncorrected half-reading method underestimates positive $R_{\mathrm{th}}$. In the original $12\mathrm{V}$ example, a stipulated $R_m=10\mathrm{k\Omega }$ gives $V_m=120/17\mathrm{V}\approx 7.059\mathrm{V}$ and half-reading load $R_L=20/17\mathrm{k\Omega }\approx 1.176\mathrm{k\Omega }$, instead of $\frac{4}{3}\mathrm{k\Omega }$. These are model predictions for a deliberately loading meter, not the specification or observed readout of the source experiment's instrument.

Finite readings can also estimate the terminal line without a physical short. For two distinct load settings with corrected delivered electric currents $I_1,I_2$ and terminal voltages $V_1,V_2$,If the voltmeter remains across each electrical load, the delivered electric current is $I_k=V_k(1/R_{L,k}+1/R_m)$, including the meter branch. Omitting it biases the inferred slope.

Useful additional controls are to measure each resistor separately before assembly, verify terminal polarity, repeat an earlier load setting to detect supply drift, and compare a third setting with the inferred Thévenin terminal response. These are proposed controls, not reported outcomes of the source experiment. Closely spaced electric currents make the slope estimate sensitive to voltage resolution and measurement error; widely separated settings may trigger source current limiting or Joule heating that changes electrical resistance. Lead and contact electrical resistance, meter input uncertainty and rounding limit the inference. An ohmmeter injects its own test electric current; its reading on a powered network is not the suppressed-network ratio. Measurements with independent sources physically removed do not characterize an active dependent electrical source whose control or supply has also been disconnected.

To determine the Thévenin resistance, remove the external electrical load and set each independent electrical source to zero in the circuit model. A zero ideal voltage source enforces zero electric potential difference, so replace it by a short connection. A zero ideal current source enforces zero electric current, so replace it by an open connection. Retain every resistor, including any explicitly modeled supply electrical resistance. Turning an actual power supply's front-panel control to zero is not evidence that it physically becomes an ideal wire.

Retain every dependent electrical source and its controlling variable: its response to an electrical test source is part of the network's electrical resistance. Impose terminal voltage $v_t$ with an electrical test source and calculate $j_t$ entering the suppressed network at $a$. For a nonzero test yielding a finite ratio,Alternatively inject a known entering electric current and solve for $v_t$. The test electric current is opposite to the delivered-current convention of the original Thévenin terminal response. ^[ref]

For an original illustration, put a positive resistor $R$ across $a,b$ and a dependent electrical source in parallel that injects electric current $g{v}_t$ from $b$ toward $a$, where $g$ has units of electrical conductance. Its control is the same terminal voltage $v_t$. Kirchhoff's current law givesWith $R=1\mathrm{k\Omega }$ and $g=0.5\mathrm{mS}$, a $1\mathrm{V}$ test draws $0.5\mathrm{mA}$, giving $R_{\mathrm{th}}=2\mathrm{k\Omega }$. Removing the dependent electrical source would falsely give $1\mathrm{k\Omega }$. With $g=2\mathrm{mS}$, the model instead has $R_{\mathrm{th}}=-1\mathrm{k\Omega }$; with $g=1\mathrm{mS}$ it draws zero test electric current for any imposed voltage and has no finite ratio. An active physical realization requires a supply and has a limited operating range.

For a finite nonzero $R_{\mathrm{th}}$ and a well-defined short-circuit response, $I_{\mathrm{sc}}=V_{\mathrm{oc}}/R_{\mathrm{th}}$. Thus $R_{\mathrm{th}}=V_{\mathrm{oc}}/I_{\mathrm{sc}}$ is usable when $I_{\mathrm{sc}}\ne 0$. A source-free positive resistor has both $V_{\mathrm{oc}}=0$ and $I_{\mathrm{sc}}=0$, so this quotient cannot reveal its electrical resistance; an electrical test source can.

Thévenin's theorem replaces the terminal behavior of a linear resistive circuit by an ideal voltage source in series with a Thévenin resistance, whenever its terminal electric potential difference is a well-defined affine function of the delivered electric current. Fix terminals $a,b$, write $V=\varphi (a)-\varphi (b)$, and take $I$ positive out of the network at $a$ into the electrical load, returning through $b$. ThenThe open-circuit voltage $V_{\mathrm{oc}}$ is the intercept at $I=0$; $-R_{\mathrm{th}}$ is the slope. Thus two parameters predict what happens when many different electrical loads are connected. The Wheatstone bridge detector loading calculation uses precisely this replacement to find the detector reading even when the bridge's two voltage dividers interact through its detector.

For example, a $12\mathrm{V}$ ideal voltage source with $2\mathrm{k\Omega }$ above and $4\mathrm{k\Omega }$ below a voltage divider junction supplies $V_{\mathrm{oc}}=8\mathrm{V}$ and $R_{\mathrm{th}}=\frac{4}{3}\mathrm{k\Omega }$ at that junction relative to the lower terminal. Attaching a $4\mathrm{k\Omega }$ electrical load reduces its voltage to $6\mathrm{V}$ and draws $1.5\mathrm{mA}$. The equivalent explains why an unloaded divider ratio alone cannot predict a loaded output.

The equivalence concerns the chosen terminals, including the power $VI$ delivered there. It does not identify internal branch electric currents or internal Joule heating. At open circuit the original divider still draws $2\mathrm{mA}$ from its $12\mathrm{V}$ supply, while the equivalent's series branch carries no electric current. ^[ref]

For $R_1,R_2,R_L>0$ and $V_s\ne 0$, define the fractional shortfall in output magnitude from the unloaded value byThis is a loading bias relative to the unloaded target, not the uncertainty of an instrument's measurement of the connected circuit. The equality with the signed ratio also holds for negative $V_s$, because the output retains its sign. The Thévenin resistance $R_{\mathrm{th}}=R_1\parallel R_2$ summarizes the divider's susceptibility: the relevant comparison is $R_L/R_{\mathrm{th}}$, not merely a large absolute $R_L$.

Set $a=R_1/R_2>0$ and $x=R_L/R_2>0$. ThenThe derivative proves that increasing the load electrical resistance always reduces this error. For equal divider arms, $e=1/(1+2x)$: $x=1$ gives $33\frac{1}{3}\%$, $x=10$ gives $100/21\%=4.7619\dots \%$, and $x=100$ gives $100/201\%=0.497512\dots \%$. Thus a “ten times the lower resistor” rule does not guarantee one-percent agreement with the unloaded target.

A quantitative design requirement follows without a rule of thumb. For a specified $0<ϵ<1$,For equal $10\mathrm{k\Omega }$ arms, $R_{\mathrm{th}}=5\mathrm{k\Omega }$, so one-percent loading error requires $R_L\ge 495\mathrm{k\Omega }$. Reducing both divider electrical resistances reduces loading but increases unloaded supply electric current and Joule heating: $P=V_s^2/(R_1+R_2)$. The stipulated equal-arm example dissipates $5\mathrm{mW}$ in total when unloaded. This tradeoff explains why the ratio alone does not determine a useful divider.